Question

What point is symmetrical to (-1, -3) for the graph of y - 2? + 12x + 8?

Answer 1

To find the point that is symmetrical to the point (-1, -3) with respect to the line represented by the equation \(y - 2 = 12x + 8\), we first need to rewrite this equation in slope-intercept form (y = mx + b).

Starting with the given equation:

\[
y - 2 = 12x + 8
\]
We add 2 to both sides to obtain:

\[
y = 12x + 10
\]

This is the line with a slope of 12 and a y-intercept at (0, 10).

Next, we want to find the perpendicular distance from the point (-1, -3) to the line. The slope of the line is 12, so the slope of the line perpendicular to it is the negative reciprocal, which is \(-\frac{1}{12}\).

Now we can find the equation of the line that passes through the point (-1, -3) with this perpendicular slope. Using the point-slope form of the equation of a line \(y - y_1 = m(x - x_1)\):

\[
y + 3 = -\frac{1}{12}(x + 1)
\]

Expanding this, we have:

\[
y + 3 = -\frac{1}{12}x - \frac{1}{12}
\]
Subtracting 3 from both sides gives:

\[
y = -\frac{1}{12}x - \frac{1}{12} - 3
\]

To combine the constant terms:

\(-3 = -\frac{36}{12}\), so:

\[
y = -\frac{1}{12}x - \frac{1 + 36}{12} = -\frac{1}{12}x - \frac{37}{12}
\]

Now we have the equation of the perpendicular line, which we can use to find the intersection point with the original line \(y = 12x + 10\). We will set the two equations equal to each other:

\[
12x + 10 = -\frac{1}{12}x - \frac{37}{12}
\]

To eliminate the fraction, we multiply the entire equation by 12:

\[
12(12x + 10) = -x - 37
\]
That simplifies to:

\[
144x + 120 = -x - 37
\]

Adding \(x\) to both sides gives:

\[
145x + 120 = -37
\]

Subtracting 120 from both sides:

\[
145x = -37 - 120
\]

This results in:

\[
145x = -157
\]
Dividing both sides by 145 gives:

\[
x = -\frac{157}{145}
\]

Now we can substitute this value back into the equation of the original line to find \(y\):

\[
y = 12(-\frac{157}{145}) + 10
\]
Calculating this gives:

\[
y = -\frac{1884}{145} + \frac{1450}{145} = \frac{-1884 + 1450}{145} = \frac{-434}{145}
\]

The intersection point is \((- \frac{157}{145}, - \frac{434}{145})\).

Now, we find the point symmetrical to (-1, -3). The midpoint between (-1, -3) and the symmetrical point \((x', y')\) is the intersection point we just found:

\[
\left(\frac{-1 + x'}{2}, \frac{-3 + y'}{2}\right) = \left(-\frac{157}{145}, -\frac{434}{145}\right)
\]

Thus, we can set up the equations:
1. \(\frac{-1 + x'}{2} = -\frac{157}{145}\)
2. \(\frac{-3 + y'}{2} = -\frac{434}{145}\)

Solving the first equation:

\[
-1 + x' = -\frac{314}{145}
\]
\[
x' = -\frac{314}{145} + 1 = -\frac{314}{145} + \frac{145}{145} = -\frac{169}{145}
\]

Now for \(y'\):

\[
-3 + y' = -\frac{868}{145}
\]
\[
y' = -\frac{868}{145} + 3 = -\frac{868}{145} + \frac{435}{145} = -\frac{433}{145}
\]

Therefore, the point that is symmetrical to (-1, -3) with respect to the line \(y - 2 = 12x + 8\) is:

\[
\left(-\frac{169}{145}, -\frac{433}{145}\right)
\]